Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{x^2 - 7x}{x^2 - 6x - 7} \div \dfrac{x - 10}{6x + 6} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{x^2 - 7x}{x^2 - 6x - 7} \times \dfrac{6x + 6}{x - 10} $ First factor the quadratic. $a = \dfrac{x^2 - 7x}{(x + 1)(x - 7)} \times \dfrac{6x + 6}{x - 10} $ Then factor out any other terms. $a = \dfrac{x(x - 7)}{(x + 1)(x - 7)} \times \dfrac{6(x + 1)}{x - 10} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ x(x - 7) \times 6(x + 1) } { (x + 1)(x - 7) \times (x - 10) } $ $a = \dfrac{ 6x(x - 7)(x + 1)}{ (x + 1)(x - 7)(x - 10)} $ Notice that $(x - 7)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 6x(x - 7)\cancel{(x + 1)}}{ \cancel{(x + 1)}(x - 7)(x - 10)} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $a = \dfrac{ 6x\cancel{(x - 7)}\cancel{(x + 1)}}{ \cancel{(x + 1)}\cancel{(x - 7)}(x - 10)} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $a = \dfrac{6x}{x - 10} ; \space x \neq -1 ; \space x \neq 7 $